ta có $\frac{1}{\sqrt{k}}$ = $\frac{2}{\sqrt{k} + \sqrt{k}}$ > $\frac{2}{\sqrt{k+1}+\sqrt{k}}$
= $2(\sqrt{k+1}-\sqrt{k})\Rightarrow\frac{1}{\sqrt{2}}...$>17tương tự $\frac{1}{\sqrt{k}}$ < $2(\sqrt{k}-\sqrt{k-1}) \Rightarrow\frac{1}{\sqrt{2}}+...$<18